(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(y, f(x, f(a, x))) → f(f(a, f(x, a)), f(a, y))
f(x, f(x, y)) → f(f(f(x, a), a), a)
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(y, f(x, f(a, x))) → F(f(a, f(x, a)), f(a, y))
F(y, f(x, f(a, x))) → F(a, f(x, a))
F(y, f(x, f(a, x))) → F(x, a)
F(y, f(x, f(a, x))) → F(a, y)
F(x, f(x, y)) → F(f(f(x, a), a), a)
F(x, f(x, y)) → F(f(x, a), a)
F(x, f(x, y)) → F(x, a)
The TRS R consists of the following rules:
f(y, f(x, f(a, x))) → f(f(a, f(x, a)), f(a, y))
f(x, f(x, y)) → f(f(f(x, a), a), a)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(y, f(x, f(a, x))) → F(a, y)
F(y, f(x, f(a, x))) → F(f(a, f(x, a)), f(a, y))
The TRS R consists of the following rules:
f(y, f(x, f(a, x))) → f(f(a, f(x, a)), f(a, y))
f(x, f(x, y)) → f(f(f(x, a), a), a)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) SplitQDPProof (EQUIVALENT transformation)
We show in the first subproof that some pairs and rules can be removed, afterwards, we continue with the remaining DP-Problem
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(y, f(x, f(a, x))) → F(a, y)
F(y, f(x, f(a, x))) → F(f(a, f(x, a)), f(a, y))
The TRS R consists of the following rules:
f(y, f(x, f(a, x))) → f(f(a, f(x, a)), f(a, y))
f(x, f(x, y)) → f(f(f(x, a), a), a)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) SemLabProof (SOUND transformation)
We found the following model for the rules of the TRSs R and P.
Interpretation over the domain with elements from 0 to 1.
a: 1
f: 0
F: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.1-0(a., y)
F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
F.1-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-1(a., y))
F.1-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-1(a., y))
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.1-0(a., y)
F.1-0(y, f.0-0(x, f.1-0(a., x))) → F.1-1(a., y)
F.1-0(y, f.1-0(x, f.1-1(a., x))) → F.1-1(a., y)
The TRS R consists of the following rules:
f.0-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
f.0-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
f.1-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-1(a., y))
f.1-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-1(a., y))
f.0-0(x, f.0-0(x, y)) → f.0-1(f.0-1(f.0-1(x, a.), a.), a.)
f.0-0(x, f.0-1(x, y)) → f.0-1(f.0-1(f.0-1(x, a.), a.), a.)
f.1-0(x, f.1-0(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
f.1-0(x, f.1-1(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 6 less nodes.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
The TRS R consists of the following rules:
f.0-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
f.0-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
f.1-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-1(a., y))
f.1-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-1(a., y))
f.0-0(x, f.0-0(x, y)) → f.0-1(f.0-1(f.0-1(x, a.), a.), a.)
f.0-0(x, f.0-1(x, y)) → f.0-1(f.0-1(f.0-1(x, a.), a.), a.)
f.1-0(x, f.1-0(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
f.1-0(x, f.1-1(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) UsableRulesReductionPairsProof (EQUIVALENT transformation)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
No dependency pairs are removed.
The following rules are removed from R:
f.0-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
f.0-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
f.0-0(x, f.0-0(x, y)) → f.0-1(f.0-1(f.0-1(x, a.), a.), a.)
f.0-0(x, f.0-1(x, y)) → f.0-1(f.0-1(f.0-1(x, a.), a.), a.)
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(F.0-0(x1, x2)) = x1 + x2
POL(a.) = 0
POL(f.0-0(x1, x2)) = x1 + x2
POL(f.0-1(x1, x2)) = x1 + x2
POL(f.1-0(x1, x2)) = x1 + x2
POL(f.1-1(x1, x2)) = x1 + x2
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F.0-0(y, f.0-0(x, f.1-0(a., x))) → F.0-0(f.1-0(a., f.0-1(x, a.)), f.1-0(a., y))
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
The TRS R consists of the following rules:
f.1-0(x, f.1-1(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
f.1-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-1(a., y))
f.1-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-1(a., y))
f.1-0(x, f.1-0(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F.0-0(y, f.1-0(x, f.1-1(a., x))) → F.0-0(f.1-0(a., f.1-1(x, a.)), f.1-0(a., y))
The TRS R consists of the following rules:
f.1-0(x, f.1-1(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
f.1-0(y, f.0-0(x, f.1-0(a., x))) → f.0-0(f.1-0(a., f.0-1(x, a.)), f.1-1(a., y))
f.1-0(y, f.1-0(x, f.1-1(a., x))) → f.0-0(f.1-0(a., f.1-1(x, a.)), f.1-1(a., y))
f.1-0(x, f.1-0(x, y)) → f.0-1(f.0-1(f.1-1(x, a.), a.), a.)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) PisEmptyProof (SOUND transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(17) TRUE
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(y, f(x, f(a, x))) → F(f(a, f(x, a)), f(a, y))
The TRS R consists of the following rules:
f(x, f(x, y)) → f(f(f(x, a), a), a)
f(y, f(x, f(a, x))) → f(f(a, f(x, a)), f(a, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) QDPBoundsTAProof (EQUIVALENT transformation)
The DP-Problem (P, R) could be shown to be Match-(raise-)DP-Bounded [TAB_NONLEFTLINEAR] by 0 for the Rule:
F(y, f(x, f(a, x))) → F(f(a, f(x, a)), f(a, y))
by considering the usable rules:
f(x, f(x, y)) → f(f(f(x, a), a), a)
f(y, f(x, f(a, x))) → f(f(a, f(x, a)), f(a, y))
The compatible tree automaton used to show the Match-(raise-)DP-Boundedness is represented by:
final states : [0]
transitions:
a0() → 29
#0() → 16
f0(4, 5) → 17
f0(2, 3) → 18
f0(2, 7) → 6
F0(1, 6) → 0
f0(2, 5) → 19
f0(9, 5) → 8
f0(8, 5) → 20
f0(10, 5) → 8
f0(2, 2) → 12
f0(11, 12) → 21
f0(2, 13) → 18
f0(2, 1) → 14
F0(1, 14) → 0
f0(11, 5) → 19
f0(5, 5) → 19
f0(3, 5) → 19
f0(13, 5) → 19
f0(15, 3) → 18
f0(15, 7) → 6
f0(15, 5) → 19
f0(15, 2) → 12
f0(15, 13) → 18
f0(15, 1) → 14
f0(4, 15) → 17
f0(2, 15) → 22
f0(9, 15) → 8
f0(8, 15) → 20
f0(10, 15) → 8
f0(11, 15) → 19
f0(5, 15) → 19
f0(3, 15) → 19
f0(13, 15) → 19
f0(16, 5) → 17
f0(2, 16) → 6
f0(17, 5) → 23
f0(2, 17) → 18
f0(18, 12) → 21
f0(18, 5) → 19
f0(2, 18) → 14
f0(19, 5) → 23
f0(2, 19) → 18
f0(20, 5) → 23
f0(20, 12) → 21
f0(2, 20) → 14
f0(15, 15) → 22
f0(15, 16) → 6
f0(15, 17) → 18
f0(15, 18) → 14
f0(15, 19) → 18
f0(15, 20) → 14
f0(16, 15) → 17
f0(17, 15) → 23
f0(18, 15) → 19
f0(19, 15) → 23
f0(20, 15) → 23
F0(18, 6) → 0
F0(18, 14) → 0
F0(20, 6) → 0
F0(20, 14) → 0
F0(1, 20) → 0
F0(1, 21) → 0
F0(18, 20) → 0
F0(18, 21) → 0
F0(20, 20) → 0
F0(20, 21) → 0
f0(22, 5) → 23
f0(11, 22) → 21
f0(2, 22) → 18
f0(23, 5) → 24
f0(2, 23) → 18
f0(24, 5) → 24
f0(24, 12) → 21
f0(2, 24) → 25
F0(24, 6) → 0
F0(24, 14) → 0
F0(1, 24) → 0
F0(24, 24) → 0
f0(25, 12) → 21
f0(25, 5) → 19
f0(2, 25) → 14
F0(25, 6) → 0
F0(25, 14) → 0
F0(1, 25) → 0
F0(25, 25) → 0
f0(15, 28) → 27
f0(27, 28) → 26
f0(26, 28) → 14
f0(29, 3) → 18
f0(29, 7) → 6
f0(29, 5) → 19
f0(29, 2) → 12
f0(29, 13) → 18
f0(29, 1) → 14
f0(29, 15) → 22
f0(29, 16) → 6
f0(29, 17) → 18
f0(29, 18) → 14
f0(29, 19) → 18
f0(29, 20) → 14
f0(29, 22) → 18
f0(29, 23) → 18
f0(29, 24) → 25
f0(29, 25) → 14
f0(29, 28) → 27
f0(4, 29) → 17
f0(2, 29) → 22
f0(9, 29) → 8
f0(8, 29) → 20
f0(10, 29) → 8
f0(11, 29) → 19
f0(5, 29) → 19
f0(3, 29) → 19
f0(13, 29) → 19
f0(15, 29) → 30
f0(16, 29) → 17
f0(17, 29) → 23
f0(18, 29) → 19
f0(19, 29) → 23
f0(20, 29) → 23
f0(22, 29) → 23
f0(23, 29) → 24
f0(24, 29) → 24
f0(25, 29) → 19
f0(27, 29) → 26
f0(26, 29) → 14
f0(18, 22) → 21
f0(20, 22) → 21
f0(22, 15) → 23
f0(15, 22) → 18
f0(24, 22) → 21
f0(25, 22) → 21
f0(23, 15) → 24
f0(15, 23) → 18
f0(24, 15) → 24
f0(15, 24) → 25
f0(25, 15) → 19
f0(15, 25) → 14
f0(29, 29) → 30
f0(20, 5) → 19
f0(20, 15) → 19
f0(23, 5) → 23
f0(23, 15) → 23
f0(20, 29) → 19
f0(23, 29) → 23
f0(24, 5) → 23
f0(24, 15) → 23
f0(24, 29) → 23
f0(24, 5) → 19
f0(24, 15) → 19
f0(2, 24) → 14
f0(15, 24) → 14
f0(29, 24) → 14
f0(24, 29) → 19
F0(18, 24) → 0
F0(18, 25) → 0
F0(20, 24) → 0
F0(20, 25) → 0
F0(24, 20) → 0
F0(25, 20) → 0
F0(24, 21) → 0
F0(25, 21) → 0
F0(24, 25) → 0
F0(25, 24) → 0
f0(30, 5) → 23
f0(30, 15) → 23
f0(30, 28) → 26
f0(11, 30) → 21
f0(2, 30) → 18
f0(15, 30) → 18
f0(18, 30) → 21
f0(20, 30) → 21
f0(24, 30) → 21
f0(25, 30) → 21
(20) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(x, f(x, y)) → f(f(f(x, a), a), a)
f(y, f(x, f(a, x))) → f(f(a, f(x, a)), f(a, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(22) YES